# “Easy” way to derive Black Sholes delta

I was always amazed that the Black-Scholes delta i.e. the following expression: $$\frac{\partial}{\partial S}\left[ S\cdot \Phi\left(\frac{log \left(\frac{S}{K}\right)+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)-Ke^{-rT} \cdot \Phi\left(\frac{log \left(\frac{S}{K}\right)-(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)\right]$$ Is just equal to the value of the CDF on the left: $$\Phi\left(\frac{log \left(\frac{S}{K}\right)+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)$$ And similarly for derivative with respect to $$K$$ despite the fact that $$S$$ and $$K$$ appear in the middle of both of them. Recently I've stumbled upon a paper which states that this fact follows from Euler's theorem as the expression is homogeneous of degree $$1$$ as a function of $$S$$ and $$K$$ and therefore can be written as a linear combination of its partial derivatives: $$f(S,K)=S \cdot \frac{\partial f}{\partial S}+K \cdot \frac{\partial f}{\partial K}$$ I see how this makes the above result much more reasonable but I don't see how it actually proves this fact. Imagine for example the function $$f(S,K)=S+K$$. We have: $$f(S,K)=S\cdot \frac{K}{S}+K\cdot \frac{S}{K}$$ And yet $$\frac{\partial f}{\partial S}=1 \neq \frac{K}{S}$$ as well as $$\frac{\partial f}{\partial K}=1 \neq \frac{S}{K}$$. Can this reasoning be fixed somehow, so that it is possible to compute this derivative without angaging in messy calculations?

• To be honest it is not that hard to compute the derivatives directly as shown here. This is more of a curiosity than anything else. – RRL May 18 at 7:47

## 1 Answer

You just showed that if $$f$$ is degree-one homogeneous then there exist representations of the form

$$f(S,K) = Sg(S,K) + K h(S,K),$$

where $$g$$ and $$h$$ are not the partial derivatives.

The actual statement of Euler's theorem is that a function $$f:\mathbb{R}^n\to \mathbb{R}$$ with continuous partial derivatives is homogeneous of degree $$\alpha$$ if and only if

$$\alpha f(x_1, \ldots,x_n) = \sum_{j=1}^n x_j \frac{\partial f}{\partial x_j}$$

It does not say given some representation $$\alpha f(x_1, \ldots,x_n) = \sum_{j=1}^n x_jg_j(x_1, \ldots,x_n)$$ and the fact that $$f$$ is homogeneous of degree $$\alpha$$, then it must hold that functions $$g_j$$ are the partial derivatives.

Where this helps with finding the Black-Scholes delta, though, is given that the option value is degree-one homogeneous, we know there exists a representation of the form

$$f(S,k) = S \frac{\partial f}{\partial S} + K \frac{\partial f}{\partial K} = S \Phi(d_1) +K \left(-e^{-rT}\Phi(d_2) \right)$$

By your reasoning, we either have

$$\frac{\partial f}{\partial S} = \Phi(d_1) \quad \text{or} \quad \frac{\partial f}{\partial S} = - \frac{K}{S}e^{-rT} \Phi(d_2)$$

Now consider that for fixed $$K$$ and $$S \to \infty$$, the option price behaves asymptotically as

$$f(S,K) \sim S - e^{-rT}K, \quad \frac{\partial f}{\partial S} \sim 1$$

Since, $$\Phi(d_1), \Phi(d_2) \to 1$$ as $$S \to \infty$$, only the first representation

$$\frac{\partial f}{\partial S} = \Phi(d_1) \sim 1,$$

has the correct asymptotic behavior.